a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x+y-1-x-y\right)^2\)

=1

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)

a) \(A=\dfrac{2^6\cdot9^2}{6^4\cdot8}\)

\(=\dfrac{2^6\cdot\left(3^2\right)^2}{3^4\cdot2^4\cdot2^3}\)

\(=\dfrac{2^6\cdot3^4}{3^4\cdot2^7}\)

\(=\dfrac{1}{2}\)

b) \(B=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot\left(3^2\right)^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}\)

\(=\dfrac{3}{2^2}\)

\(=\dfrac{3}{4}\)

a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

Bài 2:

a) \(=x^2-36y^2\)

b) \(=x^3-8\)

Bài 3:

a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)

b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)

b: Ta có: \(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+9\left(x+2\right)^2\)

\(=x^3-9x^2+27x-27-x^3-8+9x^2+36x+36\)

\(=53x+1\)

a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

\(=\left(x+1\right)^2\)

\(=x^2+2x+1\)